# How do you draw a phase diagram with a differential equation?

##### 1 Answer

Well, it can be sketched by knowing data such as the following:

- normal boiling point (
#T_b# at#"1 atm"# ), if applicable - normal melting point (
#T_f# at#"1 atm"# ) - triple point (
#T_"tp", P_"tp"# ) - critical point (
#T_c,P_c# ) #DeltaH_"fus"# #DeltaH_"vap"# - Density of liquid & solid

and by knowing where general phase regions are:

**EXAMPLE DATA**

Take

- sublimation temperature:
#"195 K"# (#-78.15^@ "C"# ) at#"1 atm"# - triple point:
#"216.58 K"# (#-56.57^@ "C"# ) at#"5.185 bar"# (#"5.117 atm"# ) - critical point:
#"304.23 K"# (#31.08^@ "C"# ) at#"73.825 bar"# (#"72.860 atm"# ) #DeltaH_"sub" = "25.2 kJ/mol"# at#"195 K"# (#-78.15^@ "C"# ) and#"1 atm"# - Density of liquid is greater than density of solid

#-># positive slope on the solid-liquid coexistence curve

**DERIVING THE DIFFERENTIAL EQUATIONS**

Next, consider the **chemical potential**

For example, suppose

#mu_((s)) = mu_((g))# , or#barG_((s)) = barG_((g))#

From the **Maxwell Relation** for the Gibbs' free energy, it follows that:

#dbarG_((g)) - dbarG_((s)) = dmu_((g)) - dmu_((s)) = dDeltamu_((s)->(g))#

#= -DeltabarS_"sub"dT + DeltabarV_"sub"dP = 0# where

#barS# and#barV# are the molar entropy and molar volume, respectively, of the phase change process, while#T# and#P# are temperature and pressure.

Therefore,

#DeltabarS_("sub")dT = DeltabarV_"sub"dP#

#=> (dP)/(dT) = (DeltabarS_"sub")/(DeltabarV_"sub")#

This gives the slope of the solid-gas coexistence curve. Obviously, the gas is less dense than the solid, so the slope will be positive.

Now, the *phase equilibrium*, for some phase transition at constant temperature and pressure,

#DeltaG_"tr" = DeltaH_"tr" - T_"tr"DeltaS_"tr" = 0# ,

so

#DeltaS_"tr" = (DeltaH_"tr")/T_"tr"# .

Therefore, we obtain the **Clapeyron differential equation**:

#color(blue)((dP)/(dT) = (DeltabarH_"sub")/(T_"sub"DeltabarV_"sub"))#

And analogous equations can be derived for solid/liquid and liquid/vapor equilibria. The main difference is that for the liquid/vapor equilibrium, one can use the ideal gas law to rewrite it as:

#(dP)/(dT) = (DeltabarH_"vap")/(T_bDeltabarV_"vap") = (DeltabarH_"vap"P_"vap")/(T_b cdot RT_b)#

Therefore, for the liquid/vapor phase equilibrium, we have the **Clausius-Clapeyron differential equation**:

#1/P(dP)/(dT) = color(blue)((dlnP)/(dT) = (DeltabarH_"vap")/(RT_b^2))#

**SOLVING THE DIFFERENTIAL EQUATIONS**

This isn't that bad. In fact, all we are typically looking for is whether the slope is positive or negative, and often it is positive (except for water freezing).

**Solid/Gas, Solid/Liquid Equilibria**

#(dP)/(dT) = ((+))/((+)(+)) > 0#

This equation typically isn't evaluated formally, except in small temperature ranges where it is approximately linear.

But if we were to solve it, we would need an ** exceedingly small temperature range**.

*Provided that is the case,*

#int_(P_1)^(P_2) dP = int_(T_1)^(T_2) (DeltabarH_"tr")/(DeltabarV_"tr")1/T_"tr"dT#

In this small temperature range,

#color(blue)(DeltaP_"tr" ~~ (DeltabarH_"tr")/(DeltabarV_"tr")ln(T_(tr2)/T_(tr1)))#

**Liquid/Vapor Equilibrium**

#(dlnP)/(dT) = ((+))/((+)(+)^2) > 0#

This equation could be solved as a general chemistry exercise, but not in this form.

#int_(P_1)^(P_2) dlnP = int_(T_(b1))^(T_(b2))(DeltabarH_"vap")/R 1/T^2 dT#

Assuming a small-enough temperature range that

#color(blue)(ln(P_2/P_1) = -(DeltabarH_"vap")/R[1/T_(b2) - 1/T_(b1)])#

And this is the better-known form of the **Clausius-Clapeyron equation** from general chemistry. It is used to find boiling points at new pressures, or vapor pressures at new boiling points.

**CONSTRUCTING THE PHASE DIAGRAM**

The rest is using the data one could get from using these equations on one data point to get another data point.

*I generally start by plotting the triple point and critical point, then outlining where the solid, liquid, and gas phase regions are. Then, I note which slopes are positive or negative, and sketch the curves from there.*

- sublimation temperature:
#"195 K"# (#-78.15^@ "C"# ) at#"1 atm"# - triple point:
#"216.58 K"# (#-56.57^@ "C"# ) at#"5.185 bar"# (#"5.117 atm"# ) - critical point:
#"304.23 K"# (#31.08^@ "C"# ) at#"73.825 bar"# (#"72.860 atm"# )

And as you can see, all the slopes here are positive as predicted.